Where does the problem come from? What do you know about it?
–
Gerry MyersonJul 10 '12 at 13:13

Someone give it to me ,i was curious if there is some specific method to solve like this problem.
–
FrankJul 10 '12 at 15:18

2

Given the number of +1 to @GerryMyerson's comment I guess that I am not the only person to realise that there is no totally standard/easy way to do it. Depending on where this question comes from, it might be worth continueing to search for an elementary solution or to use heavier machinery.
–
Simon MarkettJul 10 '12 at 15:34

1 Answer
1

This is an edited version of a partial answer that I posted sometime ago and subsequently deleted (not sure if resurrecting an answer is the correct thing to do after it has been up-voted and then deleted, perhaps someone will advise). If anyone can suggest where any of this can be improved, or point out any mistakes, I would be grateful.

Consider the equation
$$n(n+1)(n+2)(n+3) = m(m+1)^2(m+2)^3(m+3)^4$$
To avoid some trivialities later on, it is easy to check that there are no solutions with $m=1$ or $m=2$.

Using the fact that $n(n+1)(n+2)(n+3)$ is almost a square, we have
$$(n^2+3n+1)^2-1 = m(m+2)\times[(m+1)(m+2)(m+3)^2]^2$$
Putting $N = n^2+3n+1$ and $M = (m+1)(m+2)(m+3)^2$, this becomes
$$N^2-1 = m(m+2)M^2$$
so that
$$N^2-1 = [(m+1)^2-1]M^2.$$
Our approach now is to write this as
$$N^2 - [(m+1)^2-1]M^2 = 1,$$
which is Pell's equation, with $d = (m+1)^2-1$. In this case there is a particularly nice solution for the Pell equation, as the continued
fraction is very simple in this instance. For convenience we change notation slightly and
use $k = m+1$, so that we are looking at solutions of
$$x^2 - (k^2-1)y^2 = 1,$$
and bear in mind that for any solution $(x,y)$ we also
require $$y = (m+1)(m+2)(m+3)^2 = k(k+1)(k+2)^2.$$
So we investigate the properties of solutions of the Pell equation above by looking at the
standard continued fraction method.
We have $$\sqrt{k^2-1} = (k-1)+\cfrac{1}{1+\cfrac{1}{(2k-2)+\cfrac{1}{1+\cfrac{1}{(2k-2) + \ddots}}}}$$
which gives the first few solutions $(x_n,y_n)$ as $(1,0), (k,1), (2k^2-1,2k), \dots$.

Looking at the solutions for $y$, we see that they are generated by the recurrence relation
$$y_{n+2} = 2ky_{n+1} - y_n, \mbox{ with } y_0 = 0, y_1 = 1.$$
Recalling that we also need $y = k(k+1)(k+2)^2$, it is enough to prove that this last expression cannot be one
of the $y_n$ from the recurrence relation as follows.
Clearly, the values $y_n$ are strictly increasing, and we claim that
$$y_4 < k(k+1)(k+2)^2 < y_5$$
A bit of algebra gives $$y_4 = 8k^3-4k$$ so that
\begin{equation*}k(k+1)(k+2)^2-y_4 = k^4-3k^3+8k^2+8k = k^3(k-3)+8k^2+8k\end{equation*}
which is clearly positive for $k\geq 3$ and is easily checked to be positive for $k=1,2$.

For the right-hand inequality above, we have
$$y_5 = 16k^4-12k^2+1$$
and then
$$y_5 - k(k+1)(k+2)^2 = 16k^4-12k^2+1 - (k^4+5k^3+8k^2+4k)$$
$$ = 15k^4-5k^3-20k^2-4k+1$$
and by examining the graph (because I can't see an elegant way to do this bit) we see that this is
positive for $k\geq 2$, and we know that $k=1$ (corresponding to $m=2$) is not a solution of
the original equation.

This shows that $k(k+1)(k+2)^2$ cannot be one of the $y_n$, so that no solution of the original equation is possible.

I am sure that there ought to be a simpler solution, but I have been unable to find one.